Question

A body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity V. If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of plane is

• A. $\frac{\text{V}}{\sqrt{2}}$
• B. V
• C. 2 V
• D. $\frac{\text{V}}{2}$

Hint:

Rolling objects are always slower than sliding ones because energy is split between linear and rotational motion.

Answer 1

The Correct Option is A

Step 1: Concept
Velocity for sliding: $V = \sqrt{2gh}$. Velocity for rolling: $v = \sqrt{\frac{2gh}{1+K^2/R^2}}$.

Step 2: Meaning
For a ring, the radius of gyration $K = R$, so $K^2/R^2 = 1$.

Step 3: Analysis
$v_{roll} = \sqrt{\frac{2gh}{1+1}} = \sqrt{\frac{2gh}{2}}$.
Substituting $2gh = V^2$, we get $v_{roll} = \sqrt{V^2/2} = \frac{V}{\sqrt{2}}$.

Step 4: Conclusion
The linear velocity of the rolling ring is $V/\sqrt{2}$. Final Answer: (A)